Final Project: Bouncing Barney

by Elizabeth Gieseking

Barney is in triangular room XYZ as shown.  He starts at point A on wall   He walks in a direction parallel to  until he runs into wall   at B.  Then he turns and walks parallel to  until he runs into wall   This process continues until Barney returns to his starting point.

We will consider several questions about Barney’s path.

·       Will Barney always return to his starting point?

·       Will Barney always make the same number of turns?

·       Will Barney always travel the same distance?


In the figure below, Barney’s walk around the room is completed. Barney walks from A to B to C to D to E to F and back to A – a total of six segments.  But why did this happen and will it always be the case?

To get a better understanding of the situation, we will color code the segments.  All of the short segments on or parallel to  are green, those on or parallel to  are magenta, and those on or parallel to  are red.

Now we will consider the relationship between these parallel segments.   and  so AXBR is a parallelogram.  Opposite sides of a parallelogram are congruent, so  and   ARCD is also a parallelogram, so   We also see that YFSE and FCDS are parallelograms so   Because all of the paths are parallel to one of the sides, we end up with three possible directions of lines and a total of 15 parallelograms.  We have already looked at the green segments in the diagram above, and we see through transitivity that  or   Similarly, considering the red segments we see that , and considering the magenta segments,

What does this have to do with path lengths?  Let’s look at the sides of .    XEFA is a parallelogram, so  and EYCD is a parallelogram, so   Combining these we see that  or the sum of the distances that Barney travels parallel to  is equal to .  Similarly, and .  Combining these,

Therefore, the sum of the six distances walked by Barney is equal to the perimeter of the triangle.

Let us also consider the congruent and similar triangles in this figure.  The six small triangles,  and  are congruent by SSS.  Each of these triangles is also similar to the original triangle XZY and the interior triangle RTS.  First we will prove .   is the same angle as .   and both are cut by .  Therefore corresponding angles  and  are congruent.  If two pairs of corresponding angles are congruent, the third pair must also be congruent since the sum of angles of a triangle is always .  Thus . 

Next we will prove that .   and  are vertical angles and are thus congruent.   because opposite angles of a parallelogram are congruent.  Congruency is transitive, so .  Similarly,  and   Therefore we have three pairs of congruent angles and consequently    We already showed that  was similar to each of the small triangles, hence all of the triangles created by Barney’s walk are similar.

Let us also examine what happens as we move A toward the midpoint of

The interior triangle, RST, decreases in size as A and D move closer together.  When A and D are trisections points of , triangle XYZ is subdivided into nine congruent triangles and points R, S, and T coincide. 

When A is at the midpoint of  we see that we have three points coinciding at the midpoint of each side.  A, D, and S are at the midpoint of  B, E, and T are at the midpoint of  and C, F, and R are at the midpoint of   In this special case, triangle XYZ is divided into four congruent triangles and Barney’s path is shorter.  Now Barney only hits three walls and each segment is half of the length of the corresponding parallel wall, thus his path length is half the perimeter of the room.  If he traveled the path ABCDEFA as normal, he would trace the medial triangle twice and again travel the same distance as the perimeter of the triangle.

We also want to consider the case in which Barney’s path starts on a point along  exterior to  as shown below.

Barney stays exterior to  for his entire path.  He still travels along six segments as in the original case, but now his path is longer than the perimeter of the triangle.  However, we will show that if we consider lengths  and  to be negative, Barney’s path length will once again equal the perimeter of the triangle. 

AXEF is a parallelogram so the opposite sides are congruent and   Similarly,  and   We can also consider parallelograms ABCZ, YEDC, and FEDZ.  From these we see that ,  and .

The case in which we start at one of the vertices can be considered as either a subcase of the interior starting point or the exterior starting point.  As shown below, Barney would just walk along the perimeter of the room.

Finally, we will consider the case in which Barney does not start along any of the walls, but instead starts somewhere in the interior of the room.  We will choose to move in the direction parallel to  first.

If we continue the path, we see that Barney again ends up at point A.  This time we have one additional point G on  and consequently one additional turn since A was on the interior of the triangle.  Otherwise, Barney’s path was unchanged and the distance he traveled was again equal to the perimeter of the triangle.



Through this exploration of Bouncing Barney’s path, we have been able to examine properties of parallel lines.  By looking at the parallelograms created by these intersecting parallel lines, we were able to equate segment lengths and discover congruent triangles.  By considering congruent angles, we also saw that all of the triangles created by Barney’s path were similar.  We also discovered that regardless of Barney’s starting position, we were always able to equate his path length to the perimeter of the triangle. 







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